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The Berkeley Review MCAT Biology Part 2 PDF

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Biology IV Partnl Molecular Cell Biology Sections YI-X Section VI Structure and Function in Cells and Viruses Section VII Metabolic Components Section VIII Metabolic Pathways Section IX Genetic Information Section X Expression of Genetic Information The BERKELEY -%•• r.E»V»I-E»Ww Specializing in MCAT Preparation ERKELEY ® R • E • V • I • E • W F.O. Box 40140, Berkeley, California 94704-0140 Phone: (510) 843-8378 • (510) THE-TEST Internet: Biology A. Biological Molecules Section VI 1. Amino Acids and Proteins 2. Carbohydrates and Lipids 3. Nucleic Acids Structure and B. Eukaryotic Cells Function in 1. Cellular Reproduction a. Eukaryotic Chromosomes Cells and Viruses b. The Cell Cycle, Mitosis, and Meiosis 2. Cellular and Molecular Organization a. Biomembranes and Membrane Transport b. Nucleus, Nucleolus, and Ribosomes c. Endoplasmic Reticulum, Golgi Apparatus, Lysosomes, Peroxisomes, and Mitochondria d. Microtubules, Microfilaments, and Intermediate Filaments C. Prokaryotic Cells 1. Cellular Reproduction a. Prokaryotic Chromosomes b. The Cell Cycle 2. Cellular and Molecular Organization a. Biomembranes and Membrane Transport b. Nucleoid and Ribosomes D. Viruses 1. Architecture and Genomes 2. Infection and Genomic Expression 3. Assembly and Release of Progeny Virons Practice Passages and Answers T/te BERKELEY JJrwi^w® Specializing in MCAT Preparation Structure and Function in Cells and Viruses Top 10 Section Goals Be familiar with the basic biological molecules of cells and viruses. The four general classes ofbiological molecules are amino acids, nucleic acids, carbohydrates, and lipids. Be able to recognize their basic differences, and know the functional significance ofeach class. Know the twenty standard amino acids that are used to synthesize proteins. Beable to associate amino acid names with their structures,especially the five amino adds (Asp, Glu, Lys, Arg, andHis) thatcanhave charged polar side chains. Understand the difference between mitosis and meiosis at the cellular level. Know the different stages ofthecell cycle andhow they relate to one another. Be able to describe P the chromosome numberaridploidylevel at anystageofmitosis or meiosis. Be familiar with the different types of membrane transport processes. Have a feel for the differences betweensimplediffusion, facilitated diffusion, active transport, and bulk transport Know something about symports, antiports, uniports, andmembrane charge balance. Be familiar with the structure of membrane-bound organelles in a eukaryotic cell. Be awareofhow theyrelate to oneanother within theircellular environment. Know theirgeneral * structuralcharacteristics (e.g., single membrane, double membrane, contains DNA, etc.). Be familiar with the functions of eukaryotic organelles. Foreachof the majororganelles in a eukaryotic cell, be ableto define its function in generalterms. Is it involved in metabolism? Cellular packaging and sorting? Replication? Proteinsynthesis? Know where the different metabolic processes occur in a cell. The general metabolic reactions of glycolysis, the Krebs cycle, electrontransport, and oxidative •> phosphorylation are common toeukaryotes andprokaryotes. Where do they occur ineach cell? Know the general differences between Gram-negative and Gram-positive bacteria. Bacterial cells, for the most part, canbe divided into Gram-negative and Gram-positive cells. It is important tohavean understanding ofthestructural differences between the twocell types. Be familiar with the reproductive cycle of a virus. Thebiological diversity withinviruses is enormous. Justhavea general idea of the strategies used byenveloped andnonenveloped viruses when they infect ahost cell. Know the general differences between eukaryotes, prokaryotes, and viruses. It is important to knowthebasic differences between eukaryotic cells, prokaryotic cells, and viruses. Havean understanding ofthe size differences. Howdo theyinteract with eachother? Biology Structure &? Function in Cells &Viruses Biological Molecules Biological Molecules Amino Acids & Proteins™ Amino Acids Even though there are more than 400 naturally occurring amino acids, there are a standard set of 20 amino acids that comprise the proteins of all living species. As we will see, proteins are simply individual amino acid residues linked together in a head-to-tail fashion. All of the standard 20 amino acids are referred to as o> aminoacids (i.e., a 2-amino acid), except forproline which is referred to as an oc- imino acid. An amino acid is composed ofa basic amino group (-NH2), an acidic carboxyl group (-COOH),a hydrogen (H) atom, and a side chain which is characteristic to each amino acid. If no one particular amino acid is being discussed, the side © coo chain is usually designated as -R. The amino group of an amino acid is covalently H,N- C- H IOC attached to the a-carbon atom of the amino acid-hence the name oc-amino acid. The a-carbon atom (Figure 6-1) is the carbon atom next to the carboxyl carbon atom. Figure 6-1 An oc-amino acid. Each of the 20 standard amino acids has a particular three letter and one letter abbreviation. For example, proline can be abbreviated either as Pro or as P. Each amino acid also has a defined molecular weight (Mr) and can be placed, according to its R group (at pH 7.0), into one of three main families of amino acids. Thefirst family (Figure6-2) consists of those amino acids with nonpolar R groups. These amino acids are hydrophobic. Nonpolar R Groups © HI OII © © Hl OII © © Hl Oll © © Hl Oll © © H1 O11 © H3N— C- C- O H,N— C I — C- O H3N— CI — C- O H3N— CI — C- O H3N— CI - C- O H CH3 H3C— CH CH, H3C— CH I l I Glycine (Gly)[G] CH, H3C— CH CH2 Mr = 75 Alan Miner=(A8l9a)[A] CI H3 CI H3 Valine (Val)[V] Leucine (Leu)[L] Isoleucine (Ile)[I] Mr=117 Mr=131 Mr=131 H O H O © I ll 0 © 1 11 © © 0 © H2N— C- C- O H3N— C— C- O H,N—C-C- O H3N—C- C- O i I u CI H, CH2 Pro Mlinre=(1P1ro5)[P] CSI H3 Methionine (Met)[M] Phenylalanine (Phe)[F] Tryptophan (Trp)[W] Mr= 149 Mr=165 M. = 204 Figure 6-2 Amino acids with nonpolar side chains. Copyright © by The Berkeley Review The Berkeley Review Specializing in MCAT Preparation Biology Structure &Function in Cells 6t Viruses Biological Molecules Thesecondfamily consists of those aminoacidswith polar uncharged R groups (Figure 6-3). These amino acids are hydrophilic because they contain side chain functional groups(e.g., -OH, -SH, or -NH2) that canhydrogenbond with water, Polar Uncharged R Groups H O H H O © l 11 © I © © l 11 H3N— C— C- O H3N— C— C- O H3N— C— C- O 3 I 3 I I CH2 HO-CH CH, I OH CH3 SH Serine (Ser)[S] Threonine (Thr)[TJ Cysteine (Cys)[C] Mr=105 Mr=119 M,= 121 H O ~ H o H O © l 11 © I ll © © l ll © H,N— C- C- O H3N— C- C- O H3N- C— C~ O I J I CH2 CH2 I I 0= C- NH2 CH2 l 0= C- NH2 Asparagine (Asn)[N] Glutamine (Gln)[Q] Mr=132 Mr=146 Tyrosine (Tyr)[Y] M,= 18I Figure 6-3 Amino acids with uncharged polar side chains. The third family (Figure 6-4) consists of those amino acids with charged R groups. Note that two of the amino acids (Asp and Glu) have side chains which are carboxylic acids and three of the amino acids (Lys, Arg, and His) have side chains which are amine bases. The side chains of Asp, Glu, Lys, and Arg are highly ionized at neutral pH. The side chain of His is weakly ionized at neutral pH. We will see that the degree of ionization of an amino acid depends on its acidic and basic properties. Charged R Groups H 0 H O © l ll © © l 11 © H3N— C— C--O H3N-• c— c- 0 3 1 1 H 0 H 0 CH2 CH2 © I ll © © 1 ll © 1 1 H3N— C— C- 0 H 0 H3N— C— C- 0 CH2 CH2 3 1 © l II © 1 1 1 CH2 H3N— C— C- 0 CH2 CH2 CH2 3 1 1 1 CH2 NH r^© 1 2© 1 2© ©' l © H-N^ .N-H 0=C—0 0=C—0 NH3 H2N-• C= NH2 N^ Aspartate (Asp)[D] Glutamate (Glu)[E] Lysine (Lys)[K] Arginine (Arg)[R] Histidine (His)[H] Mr=133 Mr=147 Mr=146 M r=174 Mr=155 Figure 6-4 Amino acids with charged side chains. Copyright © by The Berkeley Review The Berkeley Review Specializing in MCAT Preparation BlOlOgy Structure & Function in Cells &Viruses Biological Molecules Although the standard 20 amino acids can exist in either the D or the L configuration (with the exception of glycine because it is non-chiral), only the L- amino acidsare found in proteins. It is not knownwhy evolution chose to incor © © coo coo porate L-amino acids into proteins instead ofD-amino acids. As anexample, the © 7 © two non-superimposable mirror images of the a-amino acid alanine are shown in H'.I1,N»-C-* H Mi C-^ NH Figure 6-5. Note that the a-carbon of alanine is chiral (there are 4 different CH, CH, substituents attached to that a-carbon). The D isomer and the L isomer of alanine Mirror are called enantiomers. The D and L isomers of the different amino acids are L-alanine D-alanine based on the absolute configuration ofD and L-glyceraldehyde. [All amino acids mentioned in these readings will be in the L-configuration, unless otherwise Figure 6-5 stated.] Non-supcrimposable mirror images. Even though D-amino acids are never found in proteins, they do exist in many organisms. For example, D-alanine and D-glutamate are found within the rigid cell walls ofsome bacteria. The presence ofthese D-amino acids helps to prevent the degradation of the bacterial cell wall by specific enzymes called proteases (more on this later). D-Valine is found in the antibiotic valinomycin, a carrier of K+ ions across membranes. Amino acids found within proteins can be modified as shown by the examples given in Figure 6-6. Two important amino acid modifications can be found in collagen, the most abundant protein inmammals. Some ofthe proline and lysine amino acid residues found within nascent molecules of collagen are hydroxylated to give 4-hydroxyproline and 5-hydroxylysine, respectively. These modified aminoacidshelp togivecollagen itshigh tensile strength. Histamine, a powerful vasodilator, is formed from the decarboxylation of the amino acid histidine. S-Adenosylmethionine is an activated form of the amino acid methionine and can methylate protein or nucleic acid substrates. « H ° NH, © I II © ~ H ° H^N—C-C-0 © i II 0 I H3N— C— C- O H CH, © l I ' CH, !i.--:-.-i! CH, i CH, ©S- 0 H-C— OH CH, C-0 © I H-Nv . N H3N— CH2 HO OH 4-Hydroxyproline 5-Hydroxylysinc Histamine S-Adenosylmethionine Figure 6-6 Modified amino acids. Acid-Base Properties Body fluids in biological systems generally have a pH range of 6.5 to 8.0. This is considered to be physiological pH. At these pH ranges in the cell the amino and carboxyl groups of the standard amino acids are ionized. In other words, the a- amino group bears a positive charge while the a-carboxyl group bears a negative charge. The dipolar or zwitterionic nature of the amino acids is due to the fact that the pKa value for the a-amino terminal is about 9.4 while the pKa for the a- carboxyl terminal is about 2.2. Since amino acids can act as either an acid or a base they are referred to as ampholytes. Table 6-1 shows the pKa values for the Copyright © by The Berkeley Review The Berkeley Review Specializing in MCAT Preparation Biology Structure Of Function in Cells &Viruses Biological Molecules a-carboxyl and a-amino groups of a selectionof the 20. standard amino acids and their ionizable side chains. Small pKa values refer to a strong acid while large pKa values refer to a weak acid. Amino Acid pK^a-COOH) pKa(a-NH3+) pl^ (side chain groups) Glycine 2.4 9.8 — Alanine 2.3 9.9 — Aspartic Acid 2.0 9.9 3.9 (P-COOH) Glutamic Acid 2.2 9.7 4.2 (Y-COOH) Histidine 1.8 9.2 6.0 Imidazole) Cysteine 1.8 10.8 8.3 (Sulfhydryl) Tyrosine 2.2 9.1 10.1 (Phenol) Lysine 2.2 9.2 10.8 (e-amino) Arginine 1.8 9.0 12.5 (Guanidino) Serine 2.2 9.2 13.0 (Hydroxyl) Table 6-1 The pK values for the ionizinggroups of a few of the standard 20 amino acids. The Henderson-Hasselbalch Equation The Henderson-Hasselbalch equation (6-1) can be used to determine the fraction of either the a-amino, a-carboxyl, or side chain groups that are ionized at a particular pH. In thisequation [Ae] is the concentration of conjugate basewhile [HAj is the concentration of conjugate acid. [A'] pH = pKa + log (6-1) [HA] Henderson-Hasselbalch Equation Let's consider the ratio of the protonated to the unprotonated a-amino group of a general amino acid at a pH of 7.0. The pKa for a typical a-amino group is about 9.4. Using the Henderson-Hasselbalch equation, we find that the ratio of the protonated a-amino group to the unprotonated a-amino group is about 102-4 or 251:1.This tells us that the predominant form of the a-amino group at a pH of 7.0 pH-PKa [A"]/HA is the protonated form. The same type of analysis for the a-carboxyl group tells -3 0.001 us that the ratio of the a-carboxylate anion to the protonated a-carboxyl group is -2 0.01 about 63,000:1. -1 0.1 0 1 101 (6-2) 7.0 = 9.4 + log [R-NH2] 2.4 =[R-NH?]=25I (6_3) [R-NHJ] [R-NH2] 1 2 100 3 1000 Similarly, at a neutral pH of 7.0 the ratio of the protonated to non-protonated Table 6-2 side chain of histidine is about 1:10. Difference between pH and pKa and the relationship to the If we know the difference between the pH and the pKa (i.e., pH - pKa), then we ratio of conjugate base to can establish a ratio (Table 6-2) between the concentration of conjugate base to conjugate acid. conjugate acid. If a particular amino acid is required at the catalytic site of an enzyme, knowing the ratio between conjugate base and conjugate acid will help determine the maximum potential activity of the enzyme. Copyright © by The Berkeley Review The Berkeley Review Specializing in WCAT Preparation Biology Structure S? Function in Cells &Viruses Biological Molecules Isoelectric Point (pi) The isoelectric point (pi) is that p H at which an amino acid (or a molecule) carries no net electric charge. By manipulating the Henderson-Hasselbalch equation wecanobtain an expression (6-4) thatstates that the isoelectric point is simply the arithmetic mean of two pKa values. , _ [pKat + pKa2] (6-4) P 2 For example, the isoelectric point for the amino acid glycine is 6.1 while the isoelectric point for the amino acid lysine is 10.0. If we were to place these two amino acids in an electric field, we would find that at any pH above their isoelectric points they would migrate toward the anode (positive electrode) while at any pH below their isoelectricpoints they would migrate toward the cathode (negative electrode). Thischaracteristic allows forseparation ofamino acids by a process called electrophoresis. Paper electrophoresis is generally used for separating mixtures that contain charged molecules which are small while gel electrophoresis is used for separating proteins and nucleic acids. Titration Curves At physiological pH glycineexists in solution as the dipolar ion. The net charge on this amino acid would be 0. However, ifwe were to add a strong acid to the dipolar solution, the carboxylate group would become protonated and the overallcharge of the amino acid would be +1.Similarly, addition of a strong base to the dipolar solution would remove a proton from the protonated a-amino group, giving the amino acid a net charge of -1. As shown in Figure 6-7, the ionization of glycine depends on the pH of the solution. PH=1 pH « 6.1 pH«14 ~ H O © ^ H ° © H O © I II © i ii e I II 0 H,N- C— C- OH =5= H3N— C- C- O H2N— C— C- O pKa H pKa H 2.4 9.8 [Net charge is+lj [Net charge is 0] [Net charge is-1] Figure 6-7 Ionization of glycine. The ionization of glycine can be followed by using a titration curve (Figure 6-8). We can start our titration by adding equivalents of base (such as NaOH) to a solution of glycine which is fully protonated. The initial pH of this solution might be in the neighborhood of 1 and the overall net charge on glycine is roughly +1.0. The first midpoint of the titration curve comes when [NH3®CH2COOH] = [NH3®CH2COOe]. The Henderson-Hasselbalchequation tellsus that at this first midpoint the pH = pKa. In other words, the pH at this point is 2.4 (which is the pKa of the first ionizable proton), and half of the solution is composed of NH3®CH2COOH while the other half of the solution is composed of NH3®CH2COOe. The overall net charge on glycine in solution at this point is about +0.5. Copyright © by The Berkeley Review The Berkeley Review Specializing in MCAT Preparation Biology Structure fie Function in Cells &Viruses Biological Molecules Addition of more base will eventually convert all of the NH3®CH2COOH to NH3®CH2COOe. At this point the first equivalence point of the solution has been reached. The overall net charge on glycine is 0. The isoelectric point has been reached and the pH is about 6.1. In order to reach the second midpoint in the titration we add more base. At the second midpoint wefind that [NH3®CH2COOe] = [NH2CH2COO®]. Again, the Henderson-Hasselbalch equation tells us that at the second midpoint the pH = pKa. The pH at this point is 9.8 and the overall net charge on glycine in solution is about -0.5. Second Midpoint 12n [NH3+CH2COO-] = [NH2CH2COO-] 0.5 1.0 [NH3+CH2COOH] Equivalence ofNaOH Added Figure 6-8 Titration of glycine. By the time the second equivalence point is reached enough base has been added to completely titrate all the ionizable protons on glycine. The dominant species in solutionis now NH2CH2COOe. Theoverallnet charge on the amino acid is -1. Buffers The ability of a solution to resist a change in its pH when either an acid or a base is added is the principle behind the buffer capacity of a solution. In Figure 6-8 note that the slope of the titration curve is much less near the first and second midpoints than near the equivalence point. For each increment of base added in the vicinity of the pKa's of glycine the pH changes very little. This means that the numerical value of the log ([Ae]/[HA1) in the Henderson-Hasselbalch equation doesnot change thatmuch. In otherwords, [Ae] approximately equals [HA] in the region of the pKa. If a weak acid is within 1 pH unit of its pKa value, it resides within a good buffering range. Copyright © by The Berkeley Review 8 The Berkeley Review Specializing in MCAT Preparation

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